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  • HDU 5303 Delicious Apples (2015多校第二场 贪心 + 枚举)

    Delicious Apples

    Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
    Total Submission(s): 321    Accepted Submission(s): 95


    Problem Description
    There are  apple trees planted along a cyclic road, which is  metres long. Your storehouse is built at position  on that cyclic road.
    The th tree is planted at position , clockwise from position . There are  delicious apple(s) on the th tree.

    You only have a basket which can contain at most  apple(s). You are to start from your storehouse, pick all the apples and carry them back to your storehouse using your basket. What is your minimum distance travelled?





    There are less than 20 huge testcases, and less than 500 small testcases.
     

    Input
    First line: , the number of testcases.
    Then  testcases follow. In each testcase:
    First line contains three integers, .
    Next  lines, each line contains .
     

    Output
    Output total distance in a line for each testcase.
     

    Sample Input
    2 10 3 2 2 2 8 2 5 1 10 4 1 2 2 8 2 5 1 0 10000
     

    Sample Output
    18 26
     
    解题思路:
    注意到,最多仅仅有一次会绕整个圈走一次。因此,先贪心的处理左半环和右半环。然后枚举绕整圈的时候从左側摘得苹果和从右側摘得苹果的数目。
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cstdlib>
    #include <vector>
    #include <queue>
    #include <stack>
    #include <cmath>
    #include <algorithm>
    #define LL long long
    using namespace std;
    const int MAXN = 100000 + 10;
    int L, N, K;
    LL x[MAXN];
    LL ld[MAXN], rd[MAXN];
    vector<LL>l, r;
    int main()
    {
        int T;
        scanf("%d", &T);
        while(T--)
        {
            scanf("%d%d%d", &L, &N, &K);
            l.clear(); r.clear();
            int pos, num, m = 0;
            for(int i=1;i<=N;i++)
            {
                scanf("%d%d", &pos, &num);
                for(int i=1;i<=num;i++)
                    x[++m] = (LL)pos;
            }
            for(int i=1;i<=m;i++)
            {
                if(2 * x[i] < L) l.push_back(x[i]);
                else r.push_back(L - x[i]);
            }
            sort(l.begin(), l.end()); sort(r.begin(), r.end());
            int lsz = l.size(), rsz = r.size();
            memset(ld, 0, sizeof(ld)); memset(rd, 0, sizeof(rd));
            for(int i=0;i<lsz;i++)
                ld[i + 1] = (i + 1 <= K ? l[i] : ld[i + 1 - K] + l[i]);
            for(int i=0;i<rsz;i++)
                rd[i + 1] = (i + 1 <= K ? r[i] : rd[i + 1 - K] + r[i]);
            LL ans = (ld[lsz] + rd[rsz]) * 2;
            for(int i=0;i<=lsz&&i<=K;i++)
            {
                int p1 = lsz - i;
                int p2 = max(0, rsz-(K-i));
                ans = min(ans, 2*(ld[p1] + rd[p2]) + L);
            }
            cout << ans << endl;
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/yutingliuyl/p/6920633.html
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